| Start | So remember we were practicing anti-derivatives.
Let's just practice one more. I'm going to put up an assignment this week, actually I'll probably put it up tomorrow, just so you're alert, ok? What did I just say? There's going to be an assignment that will probably go up tomorrow. Ok, there ya go. |
| 0:34 | Let's see if you can do that.
Alright, that's long enough. We didn't see one that looked like this, but just do the division first. Make that (8/x)-2+7x^2dx, this of course assumes that x is not 0. So I took each term and separately divided it by x, so, I get 8/x, 2x/x, which is just 2 and 7x^3/x is 7x^2. |
| 1:10 | And now you can do the integral.
You get 8ln|x|-2x+7x^3/3+C. |
| 1:32 | That annoying constant.
Now remember what I told you about the constant. The constant stands for the fact that if you took the derivative of this equation, You would get what's inside the integral, because the derivative of a constant is always 0. So it doesn't really matter what the constant is. So another type of equation that they could probably give you is something called an initial condition. So an initial condition, it doesn't have to be initial, but whatever. |
| 2:05 | Means that I give you f(0) = something.
And that helps you figure out what the constant is. Initial*0. So for example, It doesn't actually have to be 0, it can be f at any constant, we'll call it a instead of 0. Suppose I told you that f'(x) = x^2-4 and f(0)=7, find f(x). |
| 2:47 | So we have to do 2 steps. First we do the anti-derivative.
So we would say f(x) = integral of (x^2-4)dx. So remember the rule. Add one to the power, divide by the new power. |
| 3:06 | And 4, the constant, that's just 4x.
I'm getting feedback there for being too close to the blackboard. That's a C with my high quality handwriting. Ok? Now we can solve for C. Because since f(0)=7, we know that 7 = 0^3/3-4(0)+c. |
| 3:32 | These are 0, so that means the constant is 7.
Therefore, f(x)= x^3/3-4x+7. Alright let's practice a couple. And we have to use some applications. So let's do... |
| 4:52 | Ok, so there's 3 practice ones here. So you want to do the anti-derivative, and then you take the initial condition and figure out what the constant is.
|
| 5:01 | Let's take these on.
So each of these you do the same thing. You're going to do the anti-derivative, you're going to integrate, then you're going to use the information that f(1) =9, and find the constant. So the anti-derivative of 6x^2-4x+2, is 6x^3-4x^2/2+2x+C |
| 5:31 | Which simplifies to 2x^3-2x^2+2x+C.
And then I tell you that f(1) = 9. If if f(1)= 9, then 9 = 2(1)^3-2(1)^2+2(1)+C, so C = 7. |
| 6:04 | Won't give you a hard number to plug in.
Therefore, C = 7, and your equation is 2x^3-2x^2+2x+7. Ok? |
| 6:31 | Right, 9 = 2+C, so C is 7.
This is chapter 4.1 Alright, let's do the second one. Same idea. So we want to find the integral of 5e^2x. Which if we remember from last time is 5e^2x/2 +c. |
| 7:02 | Cause remember when we took the derivative, you would multiply by 2. So when you do the integral, you're gonna divide by 2.
No, don't do 5e^3x/3, cause this isn't a polynomial. This is e. Ok? So remember the integral of e^kx is just e^kx/k. |
| 7:32 | So the integral of e^2x, is just e^2x/2.
Cause remember, what's the derivative of e^x? e^x. And the integral of e^x is e^x. So it never changes. So the only thing this is doing is adjusting the constant. Alright, and then I told you f(0) = 1/2. So 1/2 = 5e^2(0)/2 + C. |
| 8:02 | And what's e^0? 1.
So this is 1/2 = 5/2 + C. C = -2. because 1/2 - 5/2 = -2. And now, you go back to the equation, and we plug in -2. We get 5e^2x/2 -2. |
| 8:33 | And that's f(x).
Ok, one more of these. Remember when you do the integral, you can always check if you got it right by taking the derivative. When you get to your answer, take the derivative and see if you get back where you started. So we'll do that in a second. |
| 9:00 | Here, we're going to want to do, first, I want to rewrite this as 4x^(-1/2).
So then the integral of 4x^(-1/2)dx becomes 4x^(1/2)/(1/2) + C. Dividing by 1/2 is the same as multiplying by 2, so this is 8*x^(1/2), or 8*(sqrt)x + C. |
| 9:42 | So if they gave you a question like this on an exam, they gotta give you something you can take the square root of. It's not interesting if I give you sqrt of 3.
x will be a perfect square. Now f(1) = -5. So -5 = 8(sqrt)1 + C. |
| 10:03 | The square root of 1 is 1, so C = -13, and therefore f(x) = 8(sqrt)x-13.
Let's check that. Let's take the derivative of that. |
| 10:30 | Well what would the derivative be? It would be 8/2(sqrt)x, the derivative of -13 would be 0,
And 8/2 is 4, so you get 4/(sqrt)x which is exactly what we had. Ok?
If we went back to 5e^2x/2, if you take the derivative, you should get back to 5e^2x at each one ok? So as I said, one way you check that you did the integration correctly is do the derivative when you get to the end and see if you get back where you started. |
| 11:08 | And as I said, as opposed to where you do multiple derivatives, we don't do multiple integrals.
Ok, so where does this stuff show up? These things actually show up in business when you take economics. Derivatives give you marginal costs, or marginal revenues. I don't know if you know what marginal cost and marginal revenue are yet, but that's the cost or revenue, let's stick with cost, |
| 11:37 | So marginal cost is the cost of making the last item that you're manufacturing.
If you're manufacturing refrigerators, and you make a million of them, it's the cost of the millionth refrigerator. So it is the last one that you're making, because remember the startup costs, you really want to know once you're making lots of refrigerators, what's the cost to make 1. |
| 12:02 | And it helps you figure out change in cost of the refrigerators, the change in cost as you go from making 1,000,000 to 1,000,001.
Marginal revenue is the revenue is the revenue you make from the last item. So imagine you're having your refrigerator manufacture plan, first you have to build the factory, then you have to buy the materials, You have to hire the workers, you have to give them training. So there's lots of costs that go in. So if you were an accountant, the cost of the first refrigerator is huge, because first you invested all those millions in the plant, |
| 12:31 | but as you start to sell more and more refrigerators, the price per refrigerator drops.
Ok? So marginal cost is a derivative. Now if you want to find out the total cost, and you have the marginal costs, you go backwards and thats what integrals can do for you. For example, I'll show you why thats true. You have some line that looks like this. It doesn't matter what that is. |
| 13:01 | And here you have the total number of refrigerators,
ok, so you're selling your refrigerators so this is just a number, and here you have your marginal cost.
|
| 13:31 | So this is the dollars per refrigerator.
Ok? That's how many dollars you spend per refrigerator, or euros, or yuan, or swiss franks, or whatever currency it doesn't really matter. So if we did the area under here, you can pretend it's a rectangle, and on this side, you've got dollars per refrigerator, |
| 14:04 | and on this side you've got refrigerators.
So when you multiply those together, what happens to the units? Well the refrigerators cancel, and you're just left with total dollars. |
| 14:35 | So an integral helps you, we're going to do more of this later in the week, helps you find the area underneath the curve.
And that area is the total dollars if this were marginal cost vs refrigerators, or marginal revenue would give you the total revenue. Theres all sorts of things, if this is miles/hour, and this is hours, you end up with miles. So if it's a physics problem, you end up with distance. |
| 15:00 | So the integral helps you find the total of something, among other things that an integral can do.
So suppose we gave you a problem, you say 'wow that's really scary how can we do this?'. A company is manufacturing those water bottles. |
| 15:35 | Which every week we seem to get another one of in this class.
If they were living I would know what's going on, but since they're inanimate, I don't know. Ok, so let's say the marginal cost of producing the xth water bottle is x^3-x |
| 16:01 | That's an unrealistic function, find the total cost if we know that the fixed cost are $7000
So this is the thing you actually might want to know about some day. Some of you might go into business yourselves.
|
| 16:30 | So if you decide you want to sell something, or if you go online and you want to sell something online,
you have an upfront cost, and you're gonna wanna figure out what your cost function is.
That's one of the joys of software and emails and all sorts of electronic things. The start up cost is pretty much the only cost. The cost of sending 1,000,000 emails vs the cost of sending 100,000 emails is essentially negligible. Which, is why you have spam. It doesn't really cost a lot of money to send 100 million emails about whatever nonsense they send you emails about. |
| 17:07 | And for us, the only cost is the time deleting it.
There's server costs, and all those kinds of things, but they're very minor. That's how in the old days if you wanted to mail a million letters about something, its very expensive to do. But an email isn't expensive at all. So all you would do is this tells you the fixed cost is $7,000. |
| 17:32 | It's a way of saying the cost of making no water bottles is $7,000. It's the startup cost.
So now we would integrate this. We get the integral of x^3-xdx We used the unfortunate letter C, But whatever, so that becomes x^4/4 - x^2/2 + C |
| 18:11 | And now C(0) = 7000. So 7000= 0^4/4-0^2/2+C, so C = 7000
It doesn't always work out that this term is the constant like it did here, so be careful.
|
| 18:32 | And that tells you therefore that your cross function is x^4/4 -x^2/2 + 7000
That's unrealistic because the cost goes up very fast. It's a crappy manufacturing process.
Cost should drop as you make more and more of them, until you reach a point where you have to have a second factory, where cost shoots up again. |
| 19:01 | So you think of a company like samsung, which make a billion different things, they could have 1 factory that can crank out 25 million samsung galaxies,
But at some point they need a second factory, and their cost shoots up again.
That's why you limit the production model, or you charge more. Alright, let's have you guys try one. If you went to the bank to borrow money, because you wanted to manufacture these things, |
| 19:33 | they would want to know if your startup costs as well as your fixed costs. What are you stuck with even if you don't sell anything.
Just to begin your process. And as I've said, this isn't just manufacturing, you want to put together stony brook, The first thing you have to do is you have to build stony brook. In my youth, they spent a long time building all these buildings. This one is newer, but I moved here in, well I was 2 years old but between 1957 and 1962 there were no classes here |
| 20:06 | because first you needed 5 years of physically building the buildings, and hiring the faculty and getting all the accreditation and all that stuff.
So there are up front costs to all sorts of things. Ok? If you want to spend a space shuttle into space, that's why we did the space shuttle because we said once we made one, which will cost a zillion dollars in research and development, once we had a space shuttle, then if you could reuse it, it was cheaper the second time you sent it up. |
| 20:32 | Not a lot cheaper, still billions of dollars, but still cheaper. As opposed to the Apollo missions, they sent the rocket up and then that was it for the rocket, they didn't use it a second time.
If you think about airplanes, depending on the airline you fly, If you fly some other countries airline, some of those planes are 40 years old or more, and they're still in use, so their upfront costs of manufacturing the plane are long gone, now it's just maintaining the plane. |
| 21:05 | Which is not cheap.
I don't know what a jet plane costs these days, but it's a small number. Certainly hundreds of millions of dollars. Alright, so fixed cost of $6500. As I said, that's the cost if you've done nothing. If you just get it going. If you've manufactured no items. So you would do the integral of (4x-8x^3)dx |
| 21:36 | Which is 4x^2/2 -8x^4/4 + C.
Which can be reduced to 2x^2 -2x^4 + C. |
| 22:01 | This is unrealistic because over time your cost will drop to 0, which doesn't make any sense.
Marginal costs can come down but 0 is hard to do. Now f(0) = 6500. So that's 6500 = 2(0)^2-2(0)^4+C, so C = 6500 So your manufacturing cost is 2x^2-2x^4+6500 |
| 22:39 | Ok?
So far so good? Let's have you guys do one more. Ok, Ooh this is a fun one. |
| 23:22 | The rate that a machine operator's efficiency changes over time (expressed as a percent) is dE/de = 30-10t, where t is in hours.
|
| 24:23 | Ok, first, ask yourself does that make sense.
Should you become less efficient over time? Or should we become more efficient over time? I don't know we'll find out. |
| 24:36 | So, maybe theres an optimal amount of time for your efficiency.
So let's see. Find E(t), given that E(2) = 72%. |
| 25:00 | And then find the efficiency at 3, 4 and 5 hours.
|
| 25:30 | Ok, so the efficiency, dE/dt, is 30-10t, so we need to integrate 30-10t.
And we get 30t-10t^2/2 + C Also known as -5t^2+30t + C so that's an upside down parabola. |
| 26:04 | Ok, that's our efficiency.
And then efficency at E(2) is 72%, so 72 = -5(2)^2-30(2)+C 72 = 40+C, C = 32. So that means you start out with an efficiency of 32. |
| 26:37 | Oh uh, yeah, I did use percentage. You guys made life to hard on yourselves and put it into decimals.
I would leave it 72, I wouldn't make it .72. The numbers become pretty awful. |
| 27:00 | Sorry about that.
So that means our efficiency equation is -5t^2-30t+32. Should have told you guys it wasn't a decimal. You're well trained though. Alright, now you just have to plug in |
| 27:34 | That's what the phrase express as a percent meant.
So E(3) = -5(9)+30(3)+32. Which is 77%. |
| 28:01 | E(4) = -5(16)+30(4)+32, which is 72%, the efficiency is dropping.
Which makes sense, too many hours in the machine, and a 5, E(5) = -5(25)+30(5)+32, which is 57. |
| 28:42 | So let's see, at 3 hours we were at our best efficiency out of those 3. And maximum efficiency, remember maximum is where the derivative is 0, and that derivative is 0 at 3hrs, so thats where maximum would be.
Ok? So you can use this information to find all sorts of fun stuff. |
| 29:05 | I have a very sleepy class today. Is it the whole daylight's saving time thing? Is that throwing you guys off?
Alright, so I have one announcement, It's possible I won't be able to have class this Wednesday, I know that makes you sad, I will know tomorrow, we are certainly having class on Friday, but I will make an announcement if we are canceling for Wednesday. |